Advanced Calculus Single Variable

5.3.2 Ratio And Root Tests

A favorite test for convergence is the ratio test. This is discussed next. It is at the other
extreme from the alternating series test, being completely oblivious to any sort of
cancellation. It only gives absolute convergence or spectacular divergence.

To verify the divergence part, note that if r > 1, then 5.7 can be turned around for some
R > 1. Showing limn→∞

|an|

= ∞. Since the nth term fails to converge to 0, it follows the
series diverges.

To see the test fails if r = 1, consider ∑n−1 and ∑n−2. The first series diverges while the
second one converges but in both cases, r = 1. (Be sure to check this last claim.)
■

The ratio test is very useful for many different examples but it is somewhat unsatisfactory
mathematically. One reason for this is the assumption that an> 0, necessitated
by the need to divide by an, and the other reason is the possibility that the limit
might not exist. The next test, called the root test removes both of these objections.
Before presenting this test, it is necessary to first prove the existence of the pth
root of any positive number. This was shown earlier in Theorem 2.11.2 but the
following lemma gives an easier treatment of this issue based on theorems about
sequences.

Lemma 5.3.8Let α > 0 be any nonnegative number and let p ∈ ℕ. Then α1∕pexists. This is the unique positive number which when raised to the pthpower gives α.

Proof: Consider the function f

(x)

≡ xp− α. Then there exists b1 such that f

(b1)

> 0
and a1 such that f

(a1)

< 0. (Why?) Now cut the interval

[a1,b1]

into two closed
intervals of equal length. Let

[a2,b2]

be one of these which has f

(a2)

f

(b2)

≤ 0.
Now do for

[a2,b2]

the same thing which was done to get

[a2,b2]

from

[a1,b1]

.
Continue this way obtaining a sequence of nested intervals

[ak,bk]

with the property
that

bk − ak = 21−k(b1 − a1).

By the nested interval theorem, there exists a unique point x in all these intervals.
Generalizing Theorem 4.4.6 slightly to include the product of of the terms of finitely many
sequences, it follows from Theorem 4.4.11 that

f (x)f (x) = lk→im∞ f (ak)f (bk) ≤ 0

Hence f

(x)

= 0. ■

Theorem 5.3.9Suppose

|an|

1∕n< R < 1 for all n sufficiently large.Then

∞∑
an converges absolutely.
n=1

If there are infinitely many values of n such that

|an|

1∕n≥ 1, then

∞∑
an diverges.
n=1

Proof: Suppose first that

|an|

1∕n< R < 1 for all n sufficiently large. Say this holds for all
n ≥ nR. Then for such n,

∘n----
|an| < R.

Therefore, for such n,

n
|an| ≤ R

and so the comparison test with a geometric series applies and gives absolute convergence as
claimed.